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LeetCode: 981. 基于时间的键值存储

1、题目描述

创建一个基于时间的键值存储类 TimeMap,它支持下面两个操作:

  1. set(string key, string value, int timestamp)

  2. 存储键 key、值 value,以及给定的时间戳 timestamp。

  3. get(string key, int timestamp)

  4. 返回先前调用 set(key, value, timestamp_prev) 所存储的值,其中 timestamp_prev <= timestamp。

  5. 如果有多个这样的值,则返回对应最大的 timestamp_prev 的那个值。
  6. 如果没有值,则返回空字符串("")。

示例 1:

输入:inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
输出:[null,null,"bar","bar",null,"bar2","bar2"]
解释:
TimeMap kv;
kv.set("foo", "bar", 1); // 存储键 "foo" 和值 "bar" 以及时间戳 timestamp = 1
kv.get("foo", 1);  // 输出 "bar"
kv.get("foo", 3); // 输出 "bar" 因为在时间戳 3 和时间戳 2 处没有对应 "foo" 的值,所以唯一的值位于时间戳 1 处(即 "bar")
kv.set("foo", "bar2", 4);
kv.get("foo", 4); // 输出 "bar2"
kv.get("foo", 5); // 输出 "bar2"

示例 2:

输入:inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
输出:[null,null,null,"","high","high","low","low"]

提示:

  • 所有的键/值字符串都是小写的。
  • 所有的键/值字符串长度都在 [1, 100] 范围内。
  • 所有 TimeMap.set 操作中的时间戳 timestamps 都是严格递增的。
  • 1 <= timestamp <= 10^7
  • TimeMap.set 和 TimeMap.get 函数在每个测试用例中将(组合)调用总计 120000 次。

2、解题思路

  • 哈希表加二分查找
from collections import defaultdict
import bisect


class TimeMap:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.data = defaultdict(list)

    def set(self, key: str, value: str, timestamp: int) -> None:
        self.data[key].append((timestamp, value))

    def get(self, key: str, timestamp: int) -> str:
        index = bisect.bisect(self.data[key], (timestamp, chr(127)))
        return self.data[key][index - 1][1] if index > 0 else ""

# Your TimeMap object will be instantiated and called as such:
# obj = TimeMap()
# obj.set(key,value,timestamp)
# param_2 = obj.get(key,timestamp)